Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $16$ years; the standard deviation is $3.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living longer than $19.7$ years.
Solution: $16$ $12.3$ $19.7$ $8.6$ $23.4$ $4.9$ $27.1$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $16$ years. We know the standard deviation is $3.7$ years, so one standard deviation below the mean is $12.3$ years and one standard deviation above the mean is $19.7$ years. Two standard deviations below the mean is $8.6$ years and two standard deviations above the mean is $23.4$ years. Three standard deviations below the mean is $4.9$ years and three standard deviations above the mean is $27.1$ years. We are interested in the probability of a gorilla living longer than $19.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the gorillas will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $12.3$ years and the other half $({16\%})$ will live longer than $19.7$ years. The probability of a particular gorilla living longer than $19.7$ years is ${16\%}$.